1.e^x+ydy=dx 2.dy/dx=y*x+y 3.y'+ycosx=e^-sinx,y(0)=3 1.e^x+ydy=dx 2.dy/dx=y*x+y 3.y'+ycosx=e^-sinx,y(0)=3
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![1.e^x+ydy=dx 2.dy/dx=y*x+y 3.y'+ycosx=e^-sinx,y(0)=3 1.e^x+ydy=dx 2.dy/dx=y*x+y 3.y'+ycosx=e^-sinx,y(0)=3](/uploads/image/z/5352242-50-2.jpg?t=1.e%5Ex%2Bydy%3Ddx+2.dy%2Fdx%3Dy%2Ax%2By+3.y%27%2Bycosx%3De%5E-sinx%2Cy%280%29%3D3+1.e%5Ex%2Bydy%3Ddx+2.dy%2Fdx%3Dy%2Ax%2By+3.y%27%2Bycosx%3De%5E-sinx%2Cy%280%29%3D3)
1.e^x+ydy=dx 2.dy/dx=y*x+y 3.y'+ycosx=e^-sinx,y(0)=3 1.e^x+ydy=dx 2.dy/dx=y*x+y 3.y'+ycosx=e^-sinx,y(0)=3
1.e^x+ydy=dx 2.dy/dx=y*x+y 3.y'+ycosx=e^-sinx,y(0)=3
1.e^x+ydy=dx
2.dy/dx=y*x+y
3.y'+ycosx=e^-sinx,y(0)=3
1.e^x+ydy=dx 2.dy/dx=y*x+y 3.y'+ycosx=e^-sinx,y(0)=3 1.e^x+ydy=dx 2.dy/dx=y*x+y 3.y'+ycosx=e^-sinx,y(0)=3
1.e^xdx+ydy=dx
(e^x-1)dx+ydy=0
通解e^x-x+(1/2)y^2=C
2.dy/dx=y*x+y
dy/dx=y(x+1)
dy/y=(x+1)dx
通解lny+(1/2)(x+1)^2+C
3.y'+ycosx=e^-sinx,y(0)=3
dy+ycosxdx=e^(-sinx)dx
dy+yd(sinx)=e^(-sinx)dx
e^sinxdy+ye^(sinx)d(sinx)=dx
e^sinxdy+yde^sinx=dx
dye^(sinx)=dx
通解ye^sinx=x+C
第一题错了吧
1.e^(x+y)dy=dx
e^ydy=e^(-x)dx
e^y+C=-e^(-x)
e^y+e^(-x)+C=0
2.dy/y=(x+1)dx
ln(y/C)=x+x^2/2
y=Ce^(x+x^2/2)
3.令y=C(x)e^(-sinx)代入得
dC(x)/dx=1
C(x)=x+c
y(0)=3 C(0)=3 c=3
y=(x+3)e^(-sinx)
我觉得第二个答题对的耶~